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What is the average "measure" inside an $N$ dimensional ball of radius $1$? Here "measure" is the length, area, volume, and so on of $N$ points in $N$-dimensional space.

For two dimensions my problem is what is the average distance between two random points in a radius $1$ disk. My friend Jay told me about this version of this problem and that the answer for this is $\frac{128}{45\pi}$ but I don't know how to verify this.

For three dimensions my problem is what is the average area of a triangle defined by three random points in a radius $1$ ball. For an $N$-dimensional ball of radius $1$, what is the average "measure" of the an $(N - 1)$-dimensional triangle defined by $N$ random points in the $N$-dimensional ball?

I also would like to know why for two dimensions it is supposedly $\frac{128}{45\pi}$. I know a little bit of calculus 1 and 2 from high school if that is needed to understand this problem.

Ѕᴀᴀᴅ
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    You should clarify that you mean a disk/ball rather than a circle/sphere. (The former allows you to choose points in the interior of the object, while the latter usually refers only to the boundary.) Here is an old post that discusses the problem in dimension $2$. – angryavian May 21 '21 at 23:25
  • @angryavian okay I will edit my question thanks. –  May 21 '21 at 23:26
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    So my friend Jay was wrong it's $\frac{128}{45\pi}$ –  May 21 '21 at 23:31
  • this unless the dimensions are 2 or 1 don’t have a nice easy answer. – BriggyT May 24 '21 at 23:18
  • Why not asking your friend Jay for the details of calculation in two dimensions? It's far from being trivial. – Ѕᴀᴀᴅ Jun 07 '21 at 06:47

1 Answers1

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$\def\d{\mathrm{d}}\def\R{\mathbb{R}}\def\x{\boldsymbol{x}}\def\0{\mathbf{0}}\def\abs#1{\left|#1\right|}\def\paren#1{\left(#1\right)}\def\C{\paren{\sum\limits_{\smash{j = 1}}^n C_j^2}^{\frac{1}{2}}}$This partial answer tries to show the complexity of the problem for general $n$ by deriving the formula for the content of an $(n - 1)$-dimensional body formed by $n$ points in the $n$-dimensional space.

For $\x_1, \cdots, \x_n \in \R^n$ with a general configuration, suppose $\x_i = (x_{i, 1}, \cdots, x_{i, n})$ for all $i$ and define$$ X = \begin{bmatrix} \x_1 \\ \vdots \\ \x_n \end{bmatrix} = \begin{bmatrix} x_{1, 1} & \cdots & x_{1, n}\\ \vdots & \ddots & \vdots\\ x_{n, 1} & \cdots & x_{n, n} \end{bmatrix}. $$ It is well-known that the content of the $n$-dimensional body formed by $\0, \x_1, \cdots, \x_n$ is $V = \dfrac{1}{n} |\det X|$, and the equation of the $(n - 1)$-dimensional hyperplane passing through $\x_1, \cdots, \x_n$ is\begin{gather*} \det\begin{bmatrix} \x - \x_1 \\ \x_2 - \x_1 \\ \vdots \\ \x_n - \x_1 \end{bmatrix} = \begin{vmatrix} x_1 - x_{1, 1} & x_2 - x_{1, 2} & \cdots & x_n - x_{1, n}\\ x_{2, 1} - x_{1, 1} & x_{2, 2} - x_{1, 2} & \cdots & x_{2, n} - x_{1, n}\\ \vdots & \vdots & \ddots & \vdots\\ x_{n, 1} - x_{1, 1} & x_{n, 2} - x_{1, 2} & \cdots & x_{n, n} - x_{1, n} \end{vmatrix} = 0. \tag{1} \end{gather*} Denoting the coefficient of $x_j$ in (1) by $C_j$ for each $j$, then the distance between $\0$ and the hyperplane is known to be$$ d = \frac{1}{\C} \abs{ \det\begin{bmatrix} \0 - \x_1 \\ \x_2 - \x_1 \\ \vdots \\ \x_n - \x_1 \end{bmatrix} } = \frac{1}{\C} \abs{ \det\begin{bmatrix} -\x_1 \\ \x_2 \\ \vdots \\ \x_n \end{bmatrix} } = \frac{|\det X|}{\C}, $$ which implies that the content of the $(n - 1)$-dimensional body formed by $\x_1, \cdots, \x_n$ is$$ m = \frac{nV}{d} = \C. $$

To simplify the expression of $C_j$'s, note that\begin{align*} C_1 &= \begin{vmatrix} x_{2, 2} - x_{1, 2} & \cdots & x_{2, n} - x_{1, n}\\ \vdots & \ddots & \vdots\\ x_{n, 2} - x_{1, 2} & \cdots & x_{n, n} - x_{1, n} \end{vmatrix}\\ &= \begin{vmatrix} 1 & x_{1, 2} & \cdots & x_{1, n}\\ 0 & x_{2, 2} - x_{1, 2} & \cdots & x_{2, n} - x_{1, n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & x_{n, 2} - x_{1, 2} & \cdots & x_{n, n} - x_{1, n} \end{vmatrix}\\ &= \begin{vmatrix} 1 & x_{1, 2} & \cdots & x_{1, n}\\ 1 & x_{2, 2} & \cdots & x_{2, n}\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_{n, 2} & \cdots & x_{n, n} \end{vmatrix} = \sum_{i = 1}^n A_{i, 1}, \end{align*} where $A_{i, j}$ is the $(i, j)$-th cofactor of $X$. It can be derived analogously (although with more complex notations) that $C_j = \sum\limits_{i = 1}^n A_{i, j}$ for all $j$, thus$$ d = \C = \paren{ \sum_{j = 1}^n \paren{ \sum_{i = 1}^n A_{i, j} }^2 }^{\frac{1}{2}}. $$

Therefore, the expectation to be computed is$$ \frac{1}{B_n^n} \mathop{\intop\cdots\intop}\limits_{\|\x_1\|, \cdots, \|\x_n\| \leqslant 1} \paren{ \sum_{j = 1}^n \paren{ \sum_{i = 1}^n A_{i, j}(\x_1, \cdots, \x_n) }^2 }^{\frac{1}{2}} \,\d\x_1\cdots\d\x_n, $$ where $B_n = \dfrac{π^{\frac{n}{2}}}{Γ\paren{ \frac{n}{2} + 1 }}$ is the content of the unit $n$-ball, and $\d\x_i = \d x_{i, 1}\cdots\d x_{i, n}$ for each $i$.

Ѕᴀᴀᴅ
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